Difference between revisions of "Simplifying Derivatives"

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<math>f(x)= {(x+3)^4 \over (x^2 + 5)^{1 \over 2} }</math>
<math>f(x)= {(x+3)^4 \over (x^2 + 5)^{1 \over 2} }</math>


The Quotient Rule <math>({f \over g})' = {(gf'-fg') \over g^2}
The Quotient Rule <math>({f \over g})' = {(gf'-fg') \over g^2}</math>
 
<math>f = {(x+3)}^4 </math>


<math>f = (x+3)^4 </math>
<math>f' = 4(x+3)^3 </math>
<math>f' = 4(x+3)^3 </math>
<math>g = (x^2 + 5)^{1 \over 2} </math>
<math>g = (x^2 + 5)^{1 \over 2} </math>
<math>g = {1 \over 2} (x^2 + 5)^{-1 \over 2} (2x) </math>


<math>(x^2 + 5)^{1 \over 2} 4(x+3)^3 </math>
<math>g' = {1 \over 2} (x^2 + 5)^{-1 \over 2} (2x) </math>
 
<math>{{(x^2 + 5)^{1 \over 2} 4(x+3)^3 + {(x+3)}^4 {1 \over 2} (x^2 + 5)^{-1 \over 2} (2x) } \over (x^2 + 5) }</math>
 
<math>{{4{(x^2 + 5)}(x+3)^3 - x{(x+3)}^4 (x^2 + 5)^{-1 \over 2} } \over (x^2 + 5) }</math>
 
<math>(x^2 + 5)^{-1 \over 2} {(x+3)}^3 [4(x^2 + 5)- x(x+3)] \over (x^2 +5)</math>
 
<math> {(x+3)}^3 [4(x^2 + 5)- x(x+3)] \over (x^2 +5) (x^2 + 5)^{1 \over 2}</math>
 
<math> {(x + 3)}^3 [4x^2 + 20 - x^2 - 3x] \over (x^2 +5) (x^2 + 5)^{1 \over 2}</math>
 
<math> {(x + 3)}^3 [3x^2 - 3x + 20] \over (x^2 + 5)^{3 \over 2}</math>
 
<math> {(x + 3)}^2 [3x^2 - 3x + 20] \over (x^2 + 5)^{1 \over 2}</math>

Latest revision as of 12:12, 10 October 2021

example 1

Product Rule


factorize

example 2

simplify

Chain rule

example 3

simplify

Product Rule

factorize



example 4

The Quotient Rule